A) \[Q=(-2\sqrt{2)}\,q\]
B) \[Q=-\frac{q}{2}\]
C) \[Q=(2\sqrt{2)}\,q\]
D) \[Q=-\frac{q}{2}\]
Correct Answer: D
Solution :
Three forces\[{{F}_{41}},{{F}_{42}}\]and\[{{F}_{43}}\]are acting on Q as shown resultant of\[{{F}_{41}}\]and \[{{F}_{43}}=\sqrt{2}{{F}_{each}}\] \[=\sqrt{2}\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{d}^{2}}}\] \[{{F}_{42}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\times Q}{{{(\sqrt{2}d)}^{2}}}\] \[\Rightarrow \] \[\frac{\sqrt{2}PQ}{{{d}^{2}}}=\frac{Q\times Q}{2{{d}^{2}}}\] \[\Rightarrow \] \[\sqrt{2}\times q=\frac{Q}{2}\] \[\therefore \] \[q=\frac{Q}{2\sqrt{2}}\]or\[Q=-2\sqrt{2}q\] Resultant on Q becomes zero only when q charges are of negative nature.You need to login to perform this action.
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